3.1115 \(\int x^2 (d+e x^2) (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=94 \[ \frac{1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{b x^2 \left (5 c^2 d-3 e\right )}{30 c^3}+\frac{b \left (5 c^2 d-3 e\right ) \log \left (c^2 x^2+1\right )}{30 c^5}-\frac{b e x^4}{20 c} \]

[Out]

-(b*(5*c^2*d - 3*e)*x^2)/(30*c^3) - (b*e*x^4)/(20*c) + (d*x^3*(a + b*ArcTan[c*x]))/3 + (e*x^5*(a + b*ArcTan[c*
x]))/5 + (b*(5*c^2*d - 3*e)*Log[1 + c^2*x^2])/(30*c^5)

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Rubi [A]  time = 0.139407, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {14, 4976, 446, 77} \[ \frac{1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{b x^2 \left (5 c^2 d-3 e\right )}{30 c^3}+\frac{b \left (5 c^2 d-3 e\right ) \log \left (c^2 x^2+1\right )}{30 c^5}-\frac{b e x^4}{20 c} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(d + e*x^2)*(a + b*ArcTan[c*x]),x]

[Out]

-(b*(5*c^2*d - 3*e)*x^2)/(30*c^3) - (b*e*x^4)/(20*c) + (d*x^3*(a + b*ArcTan[c*x]))/3 + (e*x^5*(a + b*ArcTan[c*
x]))/5 + (b*(5*c^2*d - 3*e)*Log[1 + c^2*x^2])/(30*c^5)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 4976

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^q, x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(1 + c^
2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && ((IGtQ[q, 0] &&  !(ILtQ[(m - 1)/2, 0] && GtQ[m +
2*q + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[q, 0] && GtQ[m + 2*q + 3, 0])) || (ILtQ[(m + 2*q + 1)/2, 0] &&
!ILtQ[(m - 1)/2, 0]))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int x^2 \left (d+e x^2\right ) \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac{1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \tan ^{-1}(c x)\right )-(b c) \int \frac{x^3 \left (5 d+3 e x^2\right )}{15+15 c^2 x^2} \, dx\\ &=\frac{1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{2} (b c) \operatorname{Subst}\left (\int \frac{x (5 d+3 e x)}{15+15 c^2 x} \, dx,x,x^2\right )\\ &=\frac{1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{2} (b c) \operatorname{Subst}\left (\int \left (\frac{5 c^2 d-3 e}{15 c^4}+\frac{e x}{5 c^2}+\frac{-5 c^2 d+3 e}{15 c^4 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac{b \left (5 c^2 d-3 e\right ) x^2}{30 c^3}-\frac{b e x^4}{20 c}+\frac{1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac{b \left (5 c^2 d-3 e\right ) \log \left (1+c^2 x^2\right )}{30 c^5}\\ \end{align*}

Mathematica [A]  time = 0.0212439, size = 119, normalized size = 1.27 \[ \frac{1}{3} a d x^3+\frac{1}{5} a e x^5+\frac{b d \log \left (c^2 x^2+1\right )}{6 c^3}+\frac{b e x^2}{10 c^3}-\frac{b e \log \left (c^2 x^2+1\right )}{10 c^5}-\frac{b d x^2}{6 c}+\frac{1}{3} b d x^3 \tan ^{-1}(c x)-\frac{b e x^4}{20 c}+\frac{1}{5} b e x^5 \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(d + e*x^2)*(a + b*ArcTan[c*x]),x]

[Out]

-(b*d*x^2)/(6*c) + (b*e*x^2)/(10*c^3) + (a*d*x^3)/3 - (b*e*x^4)/(20*c) + (a*e*x^5)/5 + (b*d*x^3*ArcTan[c*x])/3
+ (b*e*x^5*ArcTan[c*x])/5 + (b*d*Log[1 + c^2*x^2])/(6*c^3) - (b*e*Log[1 + c^2*x^2])/(10*c^5)

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Maple [A]  time = 0.037, size = 102, normalized size = 1.1 \begin{align*}{\frac{ae{x}^{5}}{5}}+{\frac{ad{x}^{3}}{3}}+{\frac{be{x}^{5}\arctan \left ( cx \right ) }{5}}+{\frac{b\arctan \left ( cx \right ) d{x}^{3}}{3}}-{\frac{bd{x}^{2}}{6\,c}}-{\frac{be{x}^{4}}{20\,c}}+{\frac{be{x}^{2}}{10\,{c}^{3}}}+{\frac{bd\ln \left ({c}^{2}{x}^{2}+1 \right ) }{6\,{c}^{3}}}-{\frac{be\ln \left ({c}^{2}{x}^{2}+1 \right ) }{10\,{c}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(e*x^2+d)*(a+b*arctan(c*x)),x)

[Out]

1/5*a*e*x^5+1/3*a*d*x^3+1/5*b*e*x^5*arctan(c*x)+1/3*b*arctan(c*x)*d*x^3-1/6*b*d*x^2/c-1/20*b*e*x^4/c+1/10*b*e*
x^2/c^3+1/6*b*d*ln(c^2*x^2+1)/c^3-1/10*b*e*ln(c^2*x^2+1)/c^5

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Maxima [A]  time = 1.083, size = 142, normalized size = 1.51 \begin{align*} \frac{1}{5} \, a e x^{5} + \frac{1}{3} \, a d x^{3} + \frac{1}{6} \,{\left (2 \, x^{3} \arctan \left (c x\right ) - c{\left (\frac{x^{2}}{c^{2}} - \frac{\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b d + \frac{1}{20} \,{\left (4 \, x^{5} \arctan \left (c x\right ) - c{\left (\frac{c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac{2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b e \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x^2+d)*(a+b*arctan(c*x)),x, algorithm="maxima")

[Out]

1/5*a*e*x^5 + 1/3*a*d*x^3 + 1/6*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*b*d + 1/20*(4*x^5*arc
tan(c*x) - c*((c^2*x^4 - 2*x^2)/c^4 + 2*log(c^2*x^2 + 1)/c^6))*b*e

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Fricas [A]  time = 1.7496, size = 244, normalized size = 2.6 \begin{align*} \frac{12 \, a c^{5} e x^{5} + 20 \, a c^{5} d x^{3} - 3 \, b c^{4} e x^{4} - 2 \,{\left (5 \, b c^{4} d - 3 \, b c^{2} e\right )} x^{2} + 4 \,{\left (3 \, b c^{5} e x^{5} + 5 \, b c^{5} d x^{3}\right )} \arctan \left (c x\right ) + 2 \,{\left (5 \, b c^{2} d - 3 \, b e\right )} \log \left (c^{2} x^{2} + 1\right )}{60 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x^2+d)*(a+b*arctan(c*x)),x, algorithm="fricas")

[Out]

1/60*(12*a*c^5*e*x^5 + 20*a*c^5*d*x^3 - 3*b*c^4*e*x^4 - 2*(5*b*c^4*d - 3*b*c^2*e)*x^2 + 4*(3*b*c^5*e*x^5 + 5*b
*c^5*d*x^3)*arctan(c*x) + 2*(5*b*c^2*d - 3*b*e)*log(c^2*x^2 + 1))/c^5

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Sympy [A]  time = 2.05625, size = 128, normalized size = 1.36 \begin{align*} \begin{cases} \frac{a d x^{3}}{3} + \frac{a e x^{5}}{5} + \frac{b d x^{3} \operatorname{atan}{\left (c x \right )}}{3} + \frac{b e x^{5} \operatorname{atan}{\left (c x \right )}}{5} - \frac{b d x^{2}}{6 c} - \frac{b e x^{4}}{20 c} + \frac{b d \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{6 c^{3}} + \frac{b e x^{2}}{10 c^{3}} - \frac{b e \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{10 c^{5}} & \text{for}\: c \neq 0 \\a \left (\frac{d x^{3}}{3} + \frac{e x^{5}}{5}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(e*x**2+d)*(a+b*atan(c*x)),x)

[Out]

Piecewise((a*d*x**3/3 + a*e*x**5/5 + b*d*x**3*atan(c*x)/3 + b*e*x**5*atan(c*x)/5 - b*d*x**2/(6*c) - b*e*x**4/(
20*c) + b*d*log(x**2 + c**(-2))/(6*c**3) + b*e*x**2/(10*c**3) - b*e*log(x**2 + c**(-2))/(10*c**5), Ne(c, 0)),
(a*(d*x**3/3 + e*x**5/5), True))

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Giac [A]  time = 1.09012, size = 162, normalized size = 1.72 \begin{align*} \frac{12 \, b c^{5} x^{5} \arctan \left (c x\right ) e + 12 \, a c^{5} x^{5} e + 20 \, b c^{5} d x^{3} \arctan \left (c x\right ) + 20 \, a c^{5} d x^{3} - 3 \, b c^{4} x^{4} e - 10 \, b c^{4} d x^{2} + 6 \, b c^{2} x^{2} e + 10 \, b c^{2} d \log \left (c^{2} x^{2} + 1\right ) - 6 \, b e \log \left (c^{2} x^{2} + 1\right )}{60 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x^2+d)*(a+b*arctan(c*x)),x, algorithm="giac")

[Out]

1/60*(12*b*c^5*x^5*arctan(c*x)*e + 12*a*c^5*x^5*e + 20*b*c^5*d*x^3*arctan(c*x) + 20*a*c^5*d*x^3 - 3*b*c^4*x^4*
e - 10*b*c^4*d*x^2 + 6*b*c^2*x^2*e + 10*b*c^2*d*log(c^2*x^2 + 1) - 6*b*e*log(c^2*x^2 + 1))/c^5